Two charged particles are separated by a distance of 1.0 cm ....

The electric force between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's law:

$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{r^2}$

For the force to be minimum, the charges must have the minimum possible magnitude. The minimum possible charge on a particle is the elementary charge $e = 1.6 \times 10^{-19} \text{ C}$.

Given the separation distance $r = 1.0 \text{ cm} = 10^{-2} \text{ m}$, substituting the values into the formula:

$F = \dfrac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{(10^{-2})^2}$

$F = \dfrac{9 \times 10^9 \times 2.56 \times 10^{-38}}{10^{-4}}$

$F = 23.04 \times 10^{-25} \text{ N}$

$F \approx 2.3 \times 10^{-24} \text{ N}$